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physics-9702 | as-a-level
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1.
Capacitance
2.
Nuclear Physics
3.
Kinematics
4.
Deformation of Solids
5.
Gravitational Fields
6.
Electricity
7.
D.C. Circuits
8.
Thermal Equilibrium
9.
Temperature Scales
10.
Magnetic Fields
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Specific Heat Capacity and Latent Heat
12.
Dynamics
13.
Medical Physics
14.
Waves
15.
The Mole
16.
Equation of State
17.
Kinetic Theory of Gases
18.
Particle Physics
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Internal Energy
20.
Forces, Density and Pressure
21.
Astronomy and Cosmology
22.
First Law of Thermodynamics
23.
Alternating Currents
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Superposition
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Simple Harmonic Oscillations
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Energy in Simple Harmonic Motion
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Quantum Physics
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Damped and Forced Oscillations, Resonance
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Work, Energy and Power
30.
Electric Fields
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Physical Quantities and Units
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Motion in a Circle
Understand that g is approximately constant for small height changes near Earth's surface
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Understand that g is Approximately Constant for Small Height Changes Near Earth's Surface
Introduction
Understanding the constancy of the acceleration due to gravity, denoted as $g$, near Earth's surface is fundamental in physics, particularly within the study of gravitational fields. This concept is pivotal for students preparing for AS & A Level Physics (9702) as it simplifies calculations and models gravitational interactions in everyday scenarios. Recognizing why $g$ remains nearly constant despite small variations in height enhances comprehension of more complex gravitational phenomena.
Key Concepts
The Acceleration Due to Gravity ($g$)
The acceleration due to gravity, commonly represented by $g$, is a measure of the gravitational force exerted by Earth on objects at or near its surface. It dictates how quickly objects accelerate downward when dropped and plays a critical role in various physical equations and principles, including Newton's law of universal gravitation and the equations of motion.
Value of $g$ Near Earth's Surface
At Earth's surface, the standard value of $g$ is approximately $9.81 \, \text{m/s}^2$. This value is determined by the mass of Earth ($M$), the gravitational constant ($G$), and the radius of Earth ($R$) through the equation:
$$
g = \frac{G M}{R^2}
$$
This formula illustrates that $g$ is inversely proportional to the square of Earth's radius, indicating that even small changes in $R$ can affect $g$, albeit slightly.
Gravitational Field Strength
The gravitational field strength at a point in space is defined as the force per unit mass experienced by a small test mass placed at that point. Near Earth's surface, the gravitational field strength is approximately the same everywhere, leading to the simplification that $g$ is constant for small height changes. This uniformity is a result of the vast radius of Earth compared to the small height variations considered.
Impact of Height on $g$
While $g$ decreases with altitude, the change is negligible for heights significantly smaller than Earth's radius. For example, at a height $h$ above the surface, the value of $g$ can be approximated by:
$$
g' = \frac{G M}{(R + h)^2} \approx g \left(1 - \frac{2h}{R}\right)
$$
Here, the term $\frac{2h}{R}$ represents the fractional change in $g$ due to the increase in radius. Given that $h \ll R$, the decrease in $g$ is minimal, justifying the assumption that $g$ remains approximately constant for small height variations.
Practical Implications of Constant $g$
Assuming $g$ is constant simplifies many physics problems, such as those involving projectile motion, free fall, and pendulum oscillations. This approximation allows students to apply basic kinematic equations without accounting for the slight variations in gravitational acceleration, facilitating a clearer understanding of underlying physical principles.
Mathematical Justification for Constant $g$
To mathematically justify the constancy of $g$ over small height changes, consider the change in gravitational acceleration with height:
$$
\Delta g = g' - g = \frac{G M}{(R + h)^2} - \frac{G M}{R^2} = \frac{G M}{R^2}\left(\frac{1}{(1 + \frac{h}{R})^2} - 1\right)
$$
For $h \ll R$, using the binomial approximation $(1 + x)^n \approx 1 + nx$ for $|x| \ll 1$, the expression simplifies to:
$$
\Delta g \approx -\frac{2 G M h}{R^3} = -2 g \frac{h}{R}
$$
Given that $R \approx 6.371 \times 10^6 \, \text{m}$ and typical height changes in scenarios like building heights or mountain elevations are much smaller, $\frac{h}{R}$ is extremely small, making $\Delta g$ negligible.
Examples Demonstrating Constant $g$ Approximation
Consider an object falling from the top of a 100-meter building. The change in gravitational acceleration can be calculated as:
$$
\Delta g = -2 g \frac{h}{R} = -2 \times 9.81 \times \frac{100}{6.371 \times 10^6} \approx -0.0031 \, \text{m/s}^2
$$
The negligible difference of approximately $0.0031 \, \text{m/s}^2$ compared to the standard $g$ value justifies treating $g$ as constant for such height changes.
Limitations of the Constant $g$ Assumption
While assuming $g$ is constant simplifies analyses, it introduces slight inaccuracies, especially in high-precision experiments or scenarios involving significant altitude changes (e.g., space travel). In such cases, the variation in $g$ with height must be accounted for to ensure accurate results.
Graphical Representation of $g$ vs. Height
Plotting $g$ against height $h$ shows a slight downward trend as height increases. However, within small height ranges (e.g., up to a few kilometers), the graph appears almost flat, visually supporting the approximation that $g$ is constant.
Conclusion on Constancy of $g$
In summary, the acceleration due to gravity remains approximately constant near Earth's surface for small height changes due to the vast radius of Earth relative to typical altitude variations considered in most physics problems. This approximation simplifies calculations and models, making it a valuable tool for students and educators in the field of physics.
Advanced Concepts
Derivation of $g$ as a Function of Height
To delve deeper, let's derive the expression for $g$ as a function of height above Earth's surface. Starting with Newton's law of universal gravitation:
$$
F = \frac{G M m}{(R + h)^2}
$$
where:
- $F$ is the gravitational force,
- $G$ is the gravitational constant,
- $M$ is the mass of Earth,
- $m$ is the mass of the object,
- $R$ is the radius of Earth,
- $h$ is the height above Earth's surface.
The gravitational field strength $g'$ at height $h$ is:
$$
g' = \frac{F}{m} = \frac{G M}{(R + h)^2}
$$
Expanding $(R + h)^2$:
$$
(R + h)^2 = R^2 \left(1 + \frac{2h}{R} + \frac{h^2}{R^2}\right)
$$
Thus,
$$
g' = \frac{G M}{R^2} \left(1 + \frac{2h}{R} + \frac{h^2}{R^2}\right)^{-1}
$$
For $h \ll R$, the binomial approximation $(1 + x)^{-1} \approx 1 - x$ applies, yielding:
$$
g' \approx \frac{G M}{R^2} \left(1 - \frac{2h}{R}\right) = g \left(1 - \frac{2h}{R}\right)
$$
This demonstrates that $g$ decreases linearly with height for small $h$ compared to $R$.
Derivation of the Binomial Approximation in Gravitational Field
The binomial approximation is crucial in simplifying the expression for $g'$ when $h \ll R$. Let's consider the general binomial expansion:
$$
(1 + x)^n \approx 1 + nx \quad \text{for} \quad |x| \ll 1
$$
Applying this to our scenario:
$$
(1 + \frac{h}{R})^{-2} \approx 1 - 2 \frac{h}{R}
$$
This approximation is valid due to the smallness of $\frac{h}{R}$. Consequently, higher-order terms like $\frac{h^2}{R^2}$ and beyond are neglected, ensuring the expression remains manageable while retaining sufficient accuracy.
Potential Energy Variation with Height
The gravitational potential energy ($U$) of an object at height $h$ is given by:
$$
U = -\frac{G M m}{R + h}
$$
For small heights, expanding the denominator using the binomial approximation:
$$
U \approx -\frac{G M m}{R} \left(1 - \frac{h}{R}\right) = U_0 \left(1 - \frac{h}{R}\right)
$$
where $U_0 = -\frac{G M m}{R}$. This linear relation indicates that potential energy increases linearly with height in the vicinity of Earth's surface, aligning with the constant $g$ approximation.
Applications in Projectile Motion
In projectile motion, assuming a constant $g$ simplifies the analysis by allowing the use of standard kinematic equations:
$$
y(t) = y_0 + v_{0y} t - \frac{1}{2} g t^2
$$
Here, $y(t)$ represents the vertical position at time $t$, $v_{0y}$ is the initial vertical velocity, and $y_0$ is the initial height. This simplification is valid for projectiles launched from and landing on Earth's surface where height variations are negligible compared to Earth's radius.
Resistance of Air and Its Effect on Constancy of $g$
While $g$ is treated as constant, air resistance can influence the motion of falling objects. Although air resistance does not alter the value of $g$, it affects the net acceleration experienced by objects, especially at higher velocities or for objects with larger surface areas. However, in many introductory physics problems, air resistance is neglected to focus solely on the effects of gravity, making the constant $g$ assumption more applicable.
Gravitational Acceleration in Different Celestial Bodies
The concept of a constant $g$ near the surface extends beyond Earth. For example, on the Moon, $g$ is approximately $1.62 \, \text{m/s}^2$. The same principles apply: for small height changes relative to the Moon's radius, $g$ can be considered constant. This universality underscores the fundamental nature of gravitational acceleration in celestial mechanics.
Experimentally Determining $g$
Historically, $g$ has been measured using pendulum experiments and free-fall methods. The pendulum method involves measuring the period of a simple pendulum:
$$
T = 2\pi \sqrt{\frac{L}{g}}
$$
where $L$ is the length of the pendulum. By rearranging, $g$ can be calculated:
$$
g = \frac{4\pi^2 L}{T^2}
$$
Assuming $g$ remains constant allows for accurate determinations of its value through precise measurements of $T$ and $L$.
Non-Uniform Gravitational Fields
In regions where the assumption of constant $g$ breaks down, such as in space or at significant heights, gravitational fields are non-uniform. Calculations in these scenarios must account for the variation of $g$ with distance, often requiring more complex models and integration techniques to determine forces and accelerations accurately.
Limitations of the Approximation in High-Energy Physics
In high-energy physics and astrophysics, where particles or objects experience significant changes in gravitational potential, the constancy of $g$ is insufficient. Relativistic effects and strong gravitational fields necessitate the use of general relativity and more advanced gravitational models to describe the behavior accurately.
Interdisciplinary Connections: Engineering and Earth Sciences
The constant $g$ approximation is not limited to physics. In engineering, it simplifies the analysis of structures and materials under gravitational loads. Similarly, in earth sciences, understanding constant gravitational acceleration assists in modeling geological processes and assessing the behavior of natural systems.
Comparison Table
| Aspect | Constant $g$ Approximation | Variable $g$ Consideration |
| Definition | Assumes gravitational acceleration remains $9.81 \, \text{m/s}^2$ near Earth's surface. | Accounts for changes in $g$ with height or position. |
| Applicability | Small height changes where $h \ll R$. | Large height changes or high-precision scenarios. |
| Simplicity | Simplifies calculations and modeling. | Requires complex calculations and models. |
| Accuracy | Sufficient for most introductory physics problems. | Necessary for advanced physics and engineering applications. |
| Examples | Projectile motion, pendulum experiments. | Spacecraft trajectory, high-altitude physics. |
Summary and Key Takeaways
- The acceleration due to gravity ($g$) is approximately constant near Earth's surface for small height changes.
- This approximation simplifies physics problems by allowing the use of basic kinematic equations.
- Mathematical derivations using the binomial approximation justify the negligible variation in $g$.
- Advanced applications require considering the variation of $g$ with height for accuracy.
- The constant $g$ concept has interdisciplinary relevance in engineering and earth sciences.
Coming Soon!
Examiner Tip
Tips
Use the mnemonic "Height High, Gravity Low" to remember that as height increases, gravitational acceleration slightly decreases.
When solving problems, always check if the height involved is much smaller than Earth's radius to justify the constant $g$ assumption.
For AP exam success, practice identifying when to apply the constant $g$ approximation versus when to consider its variation.
Did You Know
Did You Know
Despite its near-constant value, $g$ actually varies slightly depending on your location on Earth. For instance, $g$ is slightly stronger at the poles than at the equator due to Earth's oblate shape and rotation.
Moreover, astronauts experience microgravity because, in orbit, they are in a continuous free-fall state around Earth, effectively making $g$ feel almost zero despite gravity still acting on them.
Common Mistakes
Common Mistakes
Mistake: Assuming $g$ remains exactly $9.81 \, \text{m/s}^2$ regardless of height.
Correction: Recognize that $g$ decreases with height, but the change is negligible for small heights compared to Earth's radius.
Mistake: Ignoring the binomial approximation when calculating variations in $g$.
Correction: Apply the binomial approximation $(1 + x)^n \approx 1 + nx$ for $x \ll 1$ to simplify calculations.
FAQ
Why is $g$ considered constant near Earth's surface?
$g$ is treated as constant because the change in gravitational acceleration with small height variations is negligible compared to the vast radius of Earth, simplifying calculations in most practical scenarios.
How does height affect gravitational acceleration?
Gravitational acceleration decreases with height as $g$ is inversely proportional to the square of the distance from Earth's center. However, for small heights relative to Earth's radius, this decrease is minimal.
What is the binomial approximation used for in this context?
The binomial approximation simplifies the expression for $g$ at height $h$ by expanding $(1 + \frac{h}{R})^{-2}$, allowing us to approximate $g'$ as $g \left(1 - \frac{2h}{R}\right)$ when $h \ll R$.
Can the constant $g$ assumption be applied on the Moon?
Yes, the constant $g$ assumption is applicable on the Moon for small height changes relative to its radius, similar to Earth. However, the value of $g$ on the Moon is approximately $1.62 \, \text{m/s}^2$.
What are some real-world applications of assuming constant $g$?
Assuming constant $g$ is essential in calculating projectile trajectories, designing pendulums, and engineering structures where gravitational variations are insignificant.
When should the variation of $g$ with height be considered?
The variation of $g$ should be considered in high-precision experiments, aerospace engineering, and astrophysics, where significant height changes relative to Earth's radius impact gravitational calculations.
1.
Capacitance
2.
Nuclear Physics
3.
Kinematics
4.
Deformation of Solids
5.
Gravitational Fields
6.
Electricity
7.
D.C. Circuits
8.
Thermal Equilibrium
9.
Temperature Scales
10.
Magnetic Fields
11.
Specific Heat Capacity and Latent Heat
12.
Dynamics
13.
Medical Physics
14.
Waves
15.
The Mole
16.
Equation of State
17.
Kinetic Theory of Gases
18.
Particle Physics
19.
Internal Energy
20.
Forces, Density and Pressure
21.
Astronomy and Cosmology
22.
First Law of Thermodynamics
23.
Alternating Currents
24.
Superposition
25.
Simple Harmonic Oscillations
26.
Energy in Simple Harmonic Motion
27.
Quantum Physics
28.
Damped and Forced Oscillations, Resonance
29.
Work, Energy and Power
30.
Electric Fields
31.
Physical Quantities and Units
32.
Motion in a Circle
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